Don Zagier's approximation of Markov's diophantine equation

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Markov numbers are integer solutions to

x² + y² + z² = 3xyz.

The Wikipedia article on Markov numbers mentions that Don Zagier studied Markov numbers by looking the approximating equation

x² + y² + z² = 3xyz + 4/9

which is equivalent to

f(x) + f(y) = f(z)

where f(t) is defined as arccosh(3t/2). It wasn’t clear to me why the two previous equations are equivalent, so I’m writing this post to show that they are equivalent.

Examples

Before showing the equivalence of Zagier’s two equations, let’s look at an example that shows solutions to his second equation approximate solutions to Markov’s equation.

The following code verifies that (5, 13, 194) is a solution to Markov’s equation.

x, y, z = 5, 13, 194<br>assert(x**2 + y**2 + z**2 == 3*x*y*z)

With the same x and y above, let’s show that the z in Zagier’s second equation is close to the z above.

from math import cosh, acosh

f = lambda t: acosh(3*t/2)<br>g = lambda t: cosh(t)*2/3<br>z = g(f(x) + f(y))<br>print(z)

This gives z = 194.0023, which is close to the value of z in the Markov triple above.

Applying Osborn’s rule

Now suppose

f(x) + f(y) = f(z)

which expands to

arccosh(3x/2) + arccosh(3y/2)  = arccosh(3z/2).

It seems sensible to apply cosh to both sides. Is there some identity for cosh of a sum? Maybe you recall the equation for cosine of a sum:

cos(a + b) = cos(a) cos(b) − sin(a) sin(b).

Then Osborn’s rule says the corresponding hyperbolic identity is

cosh(a + b) = cosh(a) cosh(b) − sinh(a) sinh(b).

Osborn’s rule also says that the analog of the familiar identity

sin²(a) + cos²(b) = 1

is

sinh²(a) = cosh²(b) − 1.

From these two hyperbolic identities we can show that

cosh( arccosh(a) + arccosh(b) ) = ab + √(a² − 1) √(b² − 1).

Slug it out

The identity derived above is the tool we need to reduce our task to routine algebra.

If

arccosh(3x/2) + arccosh(3y/2)  = arccosh(3z/2)

then

(3x/2)  (3y/2)  + √((3x/2)² − 1) √((3y/2)² − 1) = 3z / 2

which simplifies to Zagier’s equation

x² + y² + z² = 3xyz + 4/9.

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John D. Cook, PhD

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